Question

Two identical blocks are kept on a frictionless horizontal table connected by a spring of stiffness ‘k’ and of orinignal length l0. A total charge Q is distributed on the block such that maximum elongation of

spring at equilibrium is equal to x. Value of Q is​

Answer 1

Given :

➳ Spring constant = k

➳ Original length of spring = Lo

➳ Maximum elongation of spring after putting charge Q on each block = x

➳ New length = (Lo + x)

To Find :

⟶ Value of charge Q.

SoluTion :

⇒ We know that, Electrostatic force between two like charges is repulsive and between two unlike charges is attractive.

✴ Electrostatic force :

$\bigstar\:\underline{\boxed{\bf{F_e=\dfrac{1}{4\pi \epsilon_o}\times\dfrac{Q_1\times Q_2}{r^2}}}}$

✴ Restoring spring force :

$\bigstar\:\underline{\boxed{\bf{F_s=k\Delta x}}}$

Where,

• K denotes spring constant
• Δx denoted change in length of spring

✭ At equilibrium :-

$\leadsto\tt\:F_e=F_s$

$\leadsto\tt\:\dfrac{1}{4\pi\epsilon_o}\times\dfrac{Q\times Q}{(L_o+x)^2}=k[(L_o+x)-L_o]$

$\leadsto\tt\:\dfrac{Q^2}{4\pi\epsilon_o}=kx\times (L_o-x)^2$

$\leadsto\tt\:Q^2=kx\times (L_o-x)^2\times 4\pi\epsilon_o$

$\leadsto\underline{\boxed{\bf{Q=(L_o-x)\sqrt{kx\times 4\pi\epsilon_o}}}}$

Answer 2

hope you like that that is your answer